LeeCode Combination Sum 更新动态规划法

Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

 

Because it is not confined to a specific length of the answers, so we cannot kown that before testing, which means there are not fixed end point of the backtracking.

We have to prun(剪枝） the branch(分支）, and backtrack to the last number.

It work like this:

1 We sort the numbers in non-descendant way

排序

2 Add the smallest number first, and loop to add it again, and again, until we find a solution, or the solution's value is larger than our target number.

迭代或递归求解

3 We backtrack, by pop up the number , to the last number.

回溯

4 and then we try the next number in candidates, loop to 2.

The key thoughts to modle the solution are: 

1 we use pushing number in and popping out to stand for backtracking thought.

回溯思想由push和pop体现

2 How many numbers in our solution stand for how deep level we iterate or recur.

3 we stop further iterate or recur by compare the number value in solution to the target number.

4 and further more, by knowing the way we arrange our candidates in a non-desendant way, we know if the current number won't fit for the answer, that the rest of the number in candidates won't fit too, so we don't need to try the rest. That's also the thought popular in AI, which call Heuristic Searching.

经验剪枝法

How we deal with repeated answers?

We just forward trying, don't look back the numbers in candiates.

At first, I was wondering whether we can do it in a iterative(迭代循环）way.

So I try, and it work. Below is my iterative program:

vector<vector<int> > combinationSum1(vector<int> &candidates, int target) {sort(candidates.begin(), candidates.end());vector<int> intermediate;vector<vector<int> > result;vector<int> indices;for (size_t i = 0; i < candidates.size(); ) {if(target <= candidates[i]){//找到结果，保存，但是不入栈if (target == candidates[i]){intermediate.push_back(candidates[i]);result.push_back(intermediate);intermediate.pop_back();}//注意：特殊情况-没有解或一个解，数字最小的值都大于等于targetif(intermediate.empty()) return result;i = indices.back()+1;//注意：别忘了target也要恢复上一层的数值target += intermediate.back();intermediate.pop_back();  // 本层改数值不符合规定，或者已经找到了答案indices.pop_back();while (i == candidates.size()){//别忘了这个情况//注意：都需要判断空栈的时候返回值if(intermediate.empty()) return result;i = indices.back()+1;target += intermediate.back();intermediate.pop_back();  indices.pop_back();}}else {intermediate.push_back(candidates[i]);indices.push_back(i);target -= candidates[i];}}return result;}

 But the program above is a little bit too complicate. And I think to a problem similar to this, we'd better use recursive way to solve it. So below is the recursive way.

http://discuss.leetcode.com/questions/61/combination-sum

vector<vector<int> > combinationSum(vector<int> &candidates, int target) {sort(candidates.begin(), candidates.end());vector<vector<int> > result; // 最终结果vector<int> intermediate; // 中间结果dfs(candidates, target, 0, intermediate, result);return result;}//不定深度的搜索，最好还是用递归void dfs(vector<int>& nums, int gap, int start, vector<int>& intermediate,vector<vector<int> > &result) {if (gap == 0) {  // 找到一个合法解result.push_back(intermediate);return;}for (size_t i = start; i < nums.size(); i++) { // 扩展状态//优化剪枝，如：{2,3,6,7} target=7;那么到了2,3之后7-2-3=4,nums[i]=6//这样4<6那么6和7都不用继续下面的计算了，直接返回就可以了。//这里是最好的剪枝位置了。if (gap < nums[i]) return; // 剪枝intermediate.push_back(nums[i]); // 执行扩展动作dfs(nums, gap - nums[i], i, intermediate, result);intermediate.pop_back();  // 撤销动作}}

 

2013.12.30 Update Dynamic programming algorithm. 

更新动态规划法。

 It should run faster, but we need to construct outcomes with backtracking, so actually it run just a little bit faster than using backtracking directly.

比直接使用回溯法快那么一点点。

It's a good algorithm.

I haven't seen any other website post such kind of dynamic programming approach to this problem yet.

It did have some website also use dynamic programming too, but they are not using thie algorithm like mine.

网上也有其他人写了动态规划法，但是我所看到的跟我的都不一样，有人的动态规划法比直接使用回溯慢， 我的快一点。

class Solution {public:vector<vector<int> > combinationSum(vector<int> &candidates, int target) {sort(candidates.begin(), candidates.end());vector<vector<int> > ta = genTable(candidates, target);vector<vector<int> > res;vector<int> tmp;conBtrack(ta, candidates, res, tmp, candidates.size(), target);return res;}vector<vector<int> > genTable(vector<int> &can, int tar){int n = can.size();vector<vector<int> > ta(n+1, vector<int>(tar+1));for (int i = 0; i < n+1; i++){ta[i][0] = 1;}for (int i = 1; i <= n; i++){for (int j = 1; j <= tar; j++){if (j >= can[i-1]){ta[i][j] = ta[i-1][j] + ta[i][j-can[i-1]];}else{ta[i][j] = ta[i-1][j];}}}return ta;}//很难写的递归回溯法void conBtrack(const vector<vector<int> > &ta, const vector<int>&can,vector<vector<int> > &res, vector<int> &tmp, int n, int m){while (n>0 && ta[n][m] == ta[n-1][m]) n--;if (!n || m <0) return;tmp.push_back(can[n-1]);conBtrack(ta, can, res, tmp, n, m-can[n-1]);if (m-can[n-1]==0) {res.push_back(tmp);reverse(res.back().begin(), res.back().end());}tmp.pop_back();conBtrack(ta, can, res, tmp, n-1, m);}};

2014-1-26 update

class Solution {public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {sort(candidates.begin(), candidates.end());vector<vector<int> > rs;vector<int> tmp;comb(rs, tmp, candidates, target);return rs;}void comb(vector<vector<int> > &rs, vector<int> &tmp,vector<int> &can, int tar, int index=0){if (!tmp.empty() && tar == 0){rs.push_back(tmp);}for (int i = index; i < can.size(); i++){if (tar >= can[i]){tmp.push_back(can[i]);comb(rs, tmp, can, tar-can[i], i);tmp.pop_back();}}}};